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Further Mathematics questions and answers

Further Mathematics Questions and Answers

Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.

636.

Consider the statements:

p : Musa is short

q : Musa is brilliant

Which of the following represents the statement "Musa is short but not brilliant"?

A.

pq

B.

pq

C.

pq

D.

pq

Correct answer is C

No explanation has been provided for this answer.

637.

An operation * is defined on the set, R, of real numbers by pq=p+q+2pq. If the identity element is 0, find the value of p for which the operation has no inverse.

A.

12

B.

0

C.

23

D.

2

Correct answer is A

Given the formula for p * q as: p+q+2pq and its identity element is 0, such that if, say, t is the inverse of p, then

pt=0, then p+t+2pt=0

t = \frac{-1}{1 + 2p} is the formula for the inverse of p and is undefined on R when

1 + 2p) = 0 i.e when 2p = -1; p = \frac{-1}{2}.

638.

Express \frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}} in the form p\sqrt{3} + q\sqrt{2}

A.

7\sqrt{3} - \frac{17\sqrt{2}}{3}

B.

7\sqrt{2} - \frac{17\sqrt{3}}{3}

C.

-7\sqrt{2} + \frac{17\sqrt{3}}{3}

D.

-7\sqrt{3} - \frac{17\sqrt{2}}{3}

Correct answer is B

Given \frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}},

first, we rationalise  by multiplying through with 2\sqrt{3} - 3\sqrt{2} (the inverse of the denominator).

(\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}})(\frac{2\sqrt{3} - 3\sqrt{2}}{2\sqrt{3} - 3\sqrt{2}})

= \frac{16\sqrt{3} - 24\sqrt{2} - 18\sqrt{2} + 18\sqrt{3}}{4(3) - 6\sqrt{6} + 6\sqrt{6} - 9(2)}

= \frac{34\sqrt{3} - 42\sqrt{2}}{-6} = 7\sqrt{2} - \frac{17\sqrt{3}}{3}

639.

If P = {x : -2 < x < 5} and Q = {x : -5 < x < 2} are subsets of \mu = {x : -5 \leq x \leq 5}, where x is a real number, find (P \cup Q).

A.

{x : -5 < x < 5}

B.

{x : -5 \leq x \leq 5}

C.

{x : -5 \leq x < 5}

D.

{x : -5 < x \leq 5}

Correct answer is A

P = {x : -2 < x < 5} and Q = {x:  -5 < x < 2}

(P \cup Q) = {x : -5 < x < 5}

640.

Two functions f and g are defined on the set of real numbers by f : x \to x^{2} + 1 and g : x \to x - 2. Find f o g

A.

x^{2} + 4x - 5

B.

x^{2} - 4x + 5

C.

x^{2} - 1

D.

x - 1

Correct answer is B

f(x) = x^{2} + 1  and  g(x) = x - 2

f o g = f(g(x)) = f(x - 2) = (x - 2)^{2} + 1

= x^{2} - 4x + 4 + 1 = x^{2} - 4x + 5